The molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
<h3>How to calculate molecular formula?</h3>
The molecular formula can be calculated from the empirical formula. The empirical formula of the compound is calculated as follows:
- C = 38.7% = 38.7g
- H = 9.76% = 9.76g
- O = 51.5% = 51.5g
Next, we convert the mass to moles by dividing by their atomic mass:
- C = 38.7 ÷ 12 = 3.23mol
- H = 9.76 ÷ 1 = 9.76mol
- O = 51.5÷ 16 = 3.22mol
Next, we divide by the smallest (3.22)
Hence, the empirical formula of the compound is CH3O
If the molar mass of the compound is 62g/mol;
(CH3O)n = 62
31n = 62
n = 2
(CH3O)2 = C2H6O2
Therefore, the molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
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Depends on what kind of Marijuana and Tobacco
Answer: The correct answer is it emits light.
Explanation:
When an electron is present in ground level that is the lower energy level and moves to the higher energy level, the energy is provided to the electron which means that it absorbs light. This is known as absorption.
When an electron present in the higher energy level comes back to the lower energy level, it emits radiations in the form of light. This is known as emission.
Hence, the correct answer is it emits light.
The correct answer is B. Low chemical reactivity
Answer:
The compound is most likely to be MgCl2.
Explanation:
Let the element with 2 valence electrons be X
Element with atomic number 17 is Cl.
The compound formed can be obtained as follow:
1. X will lose the 2 valence electrons to form X^2+ as shown below:
X —> X^2+ + 2e- (1)
2. Two Cl atoms will receive the 2 electrons as shown below:
2Cl + 2e- —> 2Cl- (2)
3. Combine equation 1 and 2.
X + 2Cl 2e- —> X^2+ +2Cl- + 2e-
Cancel out the electrons
X + 2Cl —> X^2+ + 2Cl-
X + 2Cl —> XCl2
Now, X is a group 2 element since it has 2 valence electrons. Therefore, X is most likely to magnesium, Mg.
Mg + 2Cl —> MgCl2.
Therefore, the compound is most likely to be magnesium chloride MgCl2